Theorem: Two triangles are congruence if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given: Two such that AB = DE, BC = EF and AC = DF.
To Prove:
Construction: Draw a line segment EG on the other side of EF such that AB = EG and . Join GF and GD.
Proof: In and , we have BC = EF [Given] AB = GE [By construction] and, [By construction] So, by SAS criterion of congruence, we have
[C.P.C.T.] Now, AB = DE and AB = GE ....(i) Similarly, AC = DF and AC = GF In , we have [From (i)] DE = GE ....(iii) In , we have DF = GF [From (ii)] ....(iv) From (iii) and (iv), we have
[Proved above] But, ....(v) Thus, in and DEF, we have AB = DE [Given] [From (v)] and, AC = DF [Given] So, by SAS criterion of congruence , we have
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Illustration: ABCD is a parallelogram , if the two diagonals are equal, find the measure of
Solution: Since ABCD is a parallelogram. Therefore, AB = CD and AD = BC [ Opposite sides of a parallelogram are equal] Thus, in and ACB, we have AD = BC [As proved above] BD = AC [Given] and, AB = AB [common] So, by SSS criterion of congruence, we have
Now, and transversal AB intersects them at A and B respectively. [Sum of the interior angles on the same side of a transversal is ] Hence, the measure of . |
In the given figure, AD is the median then ∠BAD is | |||
Right Option : D | |||
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In the figure shown above, if AB = DE, AC = DF and BC = EF, then which two triangles are congruent? | |||
Right Option : C | |||
View Explanation |
Two isosceles triangles ABC and DBC having the common base BC such that AB = AC and DB = DC. Then by which criterion? | |||
Right Option : B | |||
View Explanation |
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